You construct an ocean-going barge out of 2.50 x10^ 5 kg of iron. The shape of y

Question

You construct an ocean-going barge out of 2.50 x10^ 5 kg of iron. The shape of your barge is a rectangular bottom measuring 18.0m x15.0m and having vertical sides.

A) What is the minimum height the sides must be so that your barge will float in the ocean? (in centimeters )

B) Suppose you make the sides 75.0% higher than the minimum necessary. What is the mass of the heaviest load you could carry in the ocean without sinking? (kg)

C) If your barge goes up the Mississippi River, which is fresh water, how much extra mass can you carry without sinking? (kg)

Explanation / Answer

The mass of the iron can't exceed the mass of the water displaced
The mass of the iron is 250,000kg
The mass of the water displaced is Volume of Barge * Density of Salt Water
= 18m*15m*h*1030kg/m^3 = 270,000*h
h = 0.89 m = 89 cm

1.75*89 = 155.75cm
The new volume of water displaced is 1.55*18*15 = 418.5
The new mass of water displaced is then 418.5*1030 =431055 kg
431055-250,000 = 181055 kg

C - fresh water is less dense than salt water

if fresh water the density is 1000Kg /m^3 and so the height will be 0.925m

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